Dx - k is Hypoelliptic
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Theorem
Let $f \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a distribution.
Let $T_f$ be the distribution associated with $f$.
Let $k \in \R$.
Suppose in the distributional sense it holds that:
- $\paren {\dfrac \d {\d x} - k} T = T_f \quad (1)$
Then there is a $c \in \R$ such that:
- $T = T_{F + c \map \exp {k x} }$
where $F \in \map {C^\infty} \R$ is a solution to:
This article, or a section of it, needs explaining. In particular: General or particular? From its form it looks the latter. The source does not discuss this, but $F$ looks like a particular solution. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
- $\paren {\dfrac \d {\d x} - k} F = f$
In other words, $\dfrac \d {\d x} - k$ is a hypoelliptic operator.
Proof
By assumption $f \in \map {C^\infty} \R$.
By Solution to Linear First Order ODE with Constant Coefficients:
- $\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} F = f$
By Differentiable Function as Distribution and multiplication of distribution by a smooth function:
- $\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} T_F = T_f \quad (2)$
Subtract $(2)$ from $(1)$:
\(\ds \paren {\dfrac \d {\d x} - k} \paren {T - T_F}\) | \(=\) | \(\ds T_f - T_f\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf 0\) |
where $\mathbf 0$ is the zero distribution.
This is solved by:
- $T - T_F = T_{c \map \exp {k x} }$
or:
- $T = T_{F + c \map \exp {k x} }$
where $c \in \C$.
Finally:
- $\ds \paren {\dfrac \d {\d x} - k} \paren {F + c \map \exp {k x} } = f$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions